Convert the integrand into rational function y=(x^3+2x+1)/(x-1)^3

The integrand is
y=(x^3+2x+1)/(x-1)^3.

To convert the integrand into a
rational function, we'll use substitution technique.

We'll
substitute x-1 = t

We'll differentiate both
sides:

dx = dt

We'll express x
with respect to t:

x = t +
1

We'll write the integrand in
t:

(x^3+2x+1)/(x-1)^3 = [(t+1)^3 + 2(t+1) +
1]/t^5

We'll calculate the indefinite
integral

Int [(t+1)^3 + 2(t+1) +
1]/t^5

We'll expand the cube and we'll remove the
brackets:

Int (t^3 + 3t^2 + 3t + 1 + 2t + 2 +
1)dt/t^5

We'll combine like
terms:

Int (t^3 + 3t^2 + 5t
+ 4)dt/t^5

We'll use the additive property of
inetgrals:

Int (t^3 + 3t^2 + 5t + 4)dt/t^5=Int t^3dt/t^5 +
3Intt^2dt/t^5 + 5Int tdt/t^5 + 4 Int dt/t^5

Int (t^3 + 3t^2
+ 5t + 4)dt/t^5=Int dt/t^2+3Int dt/t^3+5Int dt/t^4 + 4 Int
dt/t^5

Int dt/t^2 = Int t^-2dt =
t^(-2+1)/(-2+1)+C

Int t^-2dt =
-1/t

3Int dt/t^3=3Int
t^-3dt

3Int t^-3dt =
-3/2t^2

5Int dt/t^4 = 5Int
t^-4dt

5Int t^-4dt = -5/3t^3

4
Int dt/t^5 = -4/4t^4

4 Int dt/t^5 =
-1/t^4

Int (t^3 + 3t^2 + 5t + 4)dt/t^5=
-1/t-3/2t^2-5/3t^3-1/t^4 + C